CALC? A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by ay = 1 2.80 m/s3)t, where the + y-direction is upward. (a) What is the height of the rocket above the surface of the earth at t = 10.0 s? (b) What is the speed of the rocket when it is 325 m above the surface of the earth?
Solution 51E Step 1 (a) The acceleration of the rocket is given by …. (1) Integrating equation (1) with respect to t we have … (2) Here c is the constant of integration. Now, we know that the rocket is starting from rest. Hence the initial velocity zero, so we have at t = 0, v(t) = 0, putting the value in above equation we have So, equation (2) becomes … (3) Again integrating the equation (3) with respect to t we get … (4) Here d is the constant of integration. Again we know that the rocket starts from the surface of the earth. Hence the displacement is zero at t = 0, using this value in equation (4), we can write that Hence the displacement at any time t is given by … (4) Now putting the value of t = 10.0 s we have At t = 10.0 s the height of the rocket is 2133 m.